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MPMC Lab Model Qeustions - EC2307

EC2307 – Microprocessors and Microcontroller Lab Model Exam Question Paper

1. Write an assembly language program for performing the Addition and Subraction     of two byte numbers using 8086 Microprocessor.

2. Write an assembly language program for performing the Multiplication and   Division of 16 bit numbers using 8086 Microprocessor.

3. Write an assembly language program to sort a given array in Ascending and     Descending order with array of length 10 using 8086 Microprocessor.

4. Write an assembly language program to find Largest and Smallest number in a given     Array using 8086 Microprocessor.

5. Write an assembly language program to move string of length FF from source to     destination using 8086 Microprocessor.

6. Write an assembly language program for Modes of operation of programable timer     8253.

7. Write an assembly language program for performing Addition and Subraction of  two   8 bit numbers using 8051 Microcontroller and store the results in memory.

8. Write an assembly language program for performing the Multiplication and Division   of two 8 bit data using 8051 and store the result in memory.

9.Write an assembly language program for performing communication between 8051     Micrcontroller  kit and PC.

10. Write an assembly language program to verify timer, interrupts and UART operations  in 8051 Microcontroller.

11. Write an assembly language program for interfacing 8279, 8259 with 8085      Microprocessor.

12. Write an assembly language program to convert Analog signal into Digital signal  using ADC interfacing using 8086 Microprocessor.

13. Write an assembly language program for Digital to Analog converter using 8086      Microprocessor.

14. Write an assembly language program to convert Digital inputs into Analog outputs  and to generate different waveforms using 8086 Microprocessor.

15. Write an assembly language program in 8086 to rotate the motor at different speeds       (using Stepper motor and DC motor).

16. Write an assembly language program in 8086 to display the rolling message       “WELCOME” in the display.

17. Write an assembly language programby interfacing 8255 with 8086 in Mode0, Mode1       and Mode2.

   

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Interfacing of Microprocessor Unit III - EC2304 MPMC Question

1.        Give the Salient features of 8253 timer module

2.        What is an sample and hold circuit

3.        List out the major function performed by CRT interface

4.        List out the four display modes of 8279 keyboard /display controller

5.        What are the commonly used ADC and DAC ‘s?

6.        Formulate the control word for setting bit PC4 high using BSR mode

7.        What is the purpose of AC flag available in Intel processors?

8.        What is key bouncing?

9.        Write down the control word format of SCON and TMOD Register

10.  Determine the control word for the following configuration of 8255:- Port A – Output

Mode of port A – Mode 1

Port B – Output

Mode of port B – Mode 0

Port C lower (pins PC0 – PC2) – Output

11.  What   is   8254?      Discuss   its   various   operating   modes.   What   are   its   areas   of applications?

12.  What is interfacing?

13.  List the operating modes of the 8255A programmable device.

14.     What is the internal frequency of the 8279? How can you drive from any available clock signal?

Part B

1. 1.   Explain the 8279 keyboard and display controller with neat sketch.
2. 2.   Describe the architecture and working of 8253 timer and also explain the various operating modes
3. 3.   With diagram, explain the operation of R-2R method of D/A Converter
4. 4.    Explain the function of CRT terminal interface

ADDRESSING MODES OF 8086

ADDRESSING MODES OF 8086

Sequential control flow instructions. 1

1.     Immediate: 2

2.     Direct 2

3.     Register 2

4.     Register Indirect 2

5.     Indexed. 2

6.     Register Relative. 3

7.     Based Indexed. 3

8.     Relative Based Indexed. 3

The control transfer instructions: 3

9.     Intra-segment direct mode: 4

10         Intra-segment Indirect Mode. 4

11. Inter-segment Direct Mode: 5

12.Inter-segment Indirect Mode: 5

Addressing mode indicates a way of locating data or operands. Depending upon the data types used in the instruction and the memory addressing modes, any instruction may belong to one or more addressing modes, or some instruction may not belong to any of the addressing modes. Thus the addressing modes describe the types of operands and the way they are accessed for executing an instruction. Here, we will present the addressing modes of the instructions depending upon their types.

 According to the flow of instruction execution, the instructions may be categorized as

(i)                 Sequential control flow instructions and

(ii)               Control transfer instructions.

Sequential control flow instructions are the instructions, which after execution, transfer control to the next instruction appearing immediately after it (in the sequence) in the program. For example, the arithmetic, logical, data transfer and processor control instructions are sequential control flow instructions.

1. Immediate: In this type of addressing, immediate data is a part of instruction, and appears in the form of successive byte or bytes.

Example: MOV AX, 0005H

In the above example, 0005H is the immediate data. The immediate data may be 8-bit or 16-bit in size.

2. Direct: In the direct addressing mode, a 16-bit memory address (offset) is directly specified in the instruction as a part of it.

Example: MOV AX, [5000H]

Here, data resides in a memory location in the data segment, whose Physical address may be computed using 5000H as the offset address and content of DS as segment address. The Physical address, here, is 10H*DS+5000H.

3. Register: In register addressing mode, the data is stored in a register and it is referred using the particular register. All the registers, except IP, may be used in this mode.

Example: MOV BX, AX.

4. Register Indirect: Sometimes, the address of the memory location, which contains data or operand, is determined in an indirect way, using the offset registers. This mode of addressing is known as register indirect mode.

In this addressing mode, the offset address of data is in either BX or SI or DI registers. The default segment is either DS or ES. The data is supposed to be available at the address pointed to by the content of any of the above registers in the default data segment.

Example: MOV AX, [BX]

Here, data is present in a memory location in DS whose offset address is in BX. The Physical address of the data is given as 10H*DS+ [BX].

5. Indexed: In this addressing mode, offset of the operand is stored in one of the index registers. DS and ES are the default segments for index registers SI and DI respectively. This mode is a special case of the above discussed register indirect addressing mode.

Example: MOV AX, [SI]

Here, data is available at an offset address stored in SI in DS. The Physical address, in this case, is computed as 10H*DS+ [SI].

6. Register Relative: In this addressing mode, the data is available at an Physical address formed by adding an 8-bit or 16-bit displacement with the content of any one of the registers BX, BP, SI and DI in the default (either DS or ES) segment. The example given before explains this mode.

       Example: MOV Ax, 50H [BX]

       Here, Physical address is given as 10H*DS+50H+ [BX].

7. Based Indexed: The Physical address of data is formed, in this addressing mode, by adding content of a base register (any one of BX or BP) to the content of an index register (any one of SI or DI). The default segment register may be ES or DS.

Example: MOV AX, [BX] [SI]

Here, BX is the base register and SI is the index register. The Physical address is computed as 10H*DS+ [BX] + [SI].

8. Relative Based Indexed: The Physical address is formed by adding an 8-bit or 16-bit displacement with the sum of contents of any one of the bases registers (BX or BP) and any one of the index registers, in a default segment.

Example: MOV AX, 50H [BX] [SI]

Here, 50H is an immediate displacement, BX is a base register and SI is an index register. The Physical address of data is computed as 160H*DS+ [BX] + [SI] + 50H.

The control transfer instructions:

 The addressing modes depend upon whether the destination location is within the same segment or a different one. It also depends upon the method of passing the destination address to the processor.

Basically, there are two addressing modes for the control transfer instructions, viz. inter-segment and intra-segment addressing modes.

If the location to which the control is to be transferred lies in a different segment other than the current one, the mode is called inter-segment mode. If the destination location lies in the same segment, the mode is called intra-segment.

Inter-segment Direct

                                         

                                                         Inter-segment

                                                                                               Inter-segment

                                                                                                Indirect

Modes for control

Transfer instructions

                                                                                               Intra-segment

                                                        Intra-segment                 Direct

                                                                 

                                                                                                     Intra-segment

                                                                                                      Indirect                      

Addressing Modes for Control Transfer Instruction

9.      Intra-segment direct mode:

In this mode, the address to which the control is to be transferred lies in the same segment in which the control transfer instruction lies and appears directly in the instruction as an immediate displacement value. In this addressing mode, the displacement is computed relative to the content of the instruction pointer IP.

The Physical address to which the control will be transferred is given by the sum of 8 or 16 bit displacement and current content of IP. In case of jump instruction, if the signed displacement (d) is of 8 bits (i.e. –128<d<+128), we term it as short jump and if it is of

16     its (i.e. –32768<+32768), it is termed as long jump.

10    Intra-segment Indirect Mode:

 In this mode, the displacement to which the control is to be transferred, is in the same segment in which the control transfer instruction lies, but it is passed to the instruction indirectly. Here, the branch address is found as the content of a register or a memory location. This addressing mode may be used in unconditional branch instructions.

11. Inter-segment Direct Mode:

In this mode, the address to which the control is to be transferred is in a different segment. This addressing mode provides a means of branching from one code segment to another code segment. Here, the CS and IP of the destination address are specified directly in the instruction.

12.Inter-segment Indirect Mode:

In this mode, the address to which the control is to be transferred lies in a different segment and it is passed to the instruction indirectly, i.e. contents of a memory block containing four bytes, i.e. IP (LSB), IP (MSB), CS (LSB) and CS (MSB) sequentially. The starting address of the memory block may be referred using any of the addressing modes, except immediate mode.

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Revision Test II- Unit II - ECE2304 Microprocessor and Microcontroller

REVISION  TEST – II

Subject: EC2304-Microprocessors and Microcontroller Class: III - ECE

1.        What is the purpose of the following comments in 8086  a) AAD b)RCL

2.        Explain the string compare instruction with syntax and suitable example

3.        What is  meant by short jump

4.        Write the ALP to multiply two 16 bit unsigned numbers to provide a 32 bit result. Assume that the two numbers are stored in CX and DX.

5.        Compare Call and Jump instruction

6.        What are the 8086 instructions used for BCD arithmetic?

7.        What is PSW?

8.        List any four program control instructions available in 8086

9.        How will carry and zero flags reflect the result of the instruction CMP BX, CX?

10.     What is the macro? .When is it used?

11.     What is segment override prefix?. Give an example

12.     What happen in 8086 when DEN=0 and DTR=1?

13.     What is the operation performed by the instruction CBW. Give an example for its use

14.     Compare Sub and CMP instruction in 8086

PART B –

1.        Write an 8086 program to convert BCD to Binary

2.        Write an  ALP to Sort the array of elements in ascending order.

3.        Write the ALP for 8086 to compute the average of n  Numbers of byte stored in the memory

4.        Explain the relative addressing mode and implied addressing mode with its syntax. Use an example

5.        Explain the assembler directives ASSUME, EQU, DW, and EVEN with suitable Examples

6.       Explain the following instruction with suitable example and syntax.LEA, LDS,LAHF, IMUL,,NEG,TEST, SAR,  JA, JBE,Jmp,